Question

The ratio between kinetic and potential energies of a body executing simple harmonic motion, when it is at a distance of $\frac{1}{N}$ of its amplitude from the mean position is

• A. $N^2+1$
• B. $\frac{1}{N^2}$
• C. $N^2$
• D. $N^2-1$

Answer 1

The correct option is D $N^2-1$

Let the amplitude of oscillation of SHM be $A$

Thus the distance of body from mean position $x=\frac{A}{N}$

Kinetic energy $K.E=\frac{1}{2}m{\omega }^2(A^2-x^2)$

$\therefore$ $K.E=\frac{1}{2}m{\omega }^2(A^2-\frac{A^2}{N^2})=\frac{1}{2}m{\omega }^2{A}^2(\frac{N^2-1}{N^2})$ ..........(1)

Potential energy $P.E=\frac{1}{2}k{x}^2$ where $k=m{\omega }^2$

$\therefore$ $P.E=\frac{1}{2}m{\omega }^2\frac{A^2}{N^2}$ .........(2)

$⟹$ $\frac{K.E}{P.E}=\frac{\frac{N^2-1}{N^2}}{\frac{1}{N^2}}=N^2-1$